Homework Help: physics

Posted by mary on Tuesday, June 9, 2009 at 6:11pm.

A 60 kg skier leaves the end of a ski-jump ramp with a velocity of 26 m/s directed 23° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 19 m/s, landing 8.3 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

I got -14330.4J, but this answer is wrong and I can't figure out why

I did KE= 09450
PE=-4880.4
ME=-14330.4

thanks

physics - Damon, Tuesday, June 9, 2009 at 7:42pm
Well I get:
initial KE = (1/2)(60)(26)^2 = 20,280 J
gain due to fall = m g h = 60(9.81)(8.3) = 4485 J
so total energy at ground if no friction = 25,165 J

actual Ke at ground = (1/2)(60)(19)^2 = 10,830 J

difference = 14,335 J which is the loss due to air drag
physics - seare, Tuesday, June 9, 2009 at 7:50pm
hi

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